By Remark REF  (b) the kernel \(\langle w_1^4 \rangle \)  of \(\pi : \operatorname{St}(C_n,{\mathbb {Z}}) \rightarrow \operatorname{Sp}_{2n}({\mathbb {Z}})\) 
is in the image of \(f\) .
Therefore, in view of the definitions of \(\operatorname{St}(C_n,{\mathbb {Z}})[2]\)  and of \(P_{2n+2}\) 
it is enough to check that the image of \(\bar{f} = \pi \circ f : B_{2n+2} \rightarrow \operatorname{Sp}_{2n}({\mathbb {Z}})\)  satisfies the following two properties:
(i) \(\bar{f}(P_{2n+2}) = \operatorname{Sp}_{2n}({\mathbb {Z}})[2]\)  and (ii) there is an isomorphism
\(\bar{f}(B_{2n+2})/\operatorname{Sp}_{2n}({\mathbb {Z}})[2] \cong {\mathfrak {S}}_{2n+2}\) .
As we observed in Part (v) of the proof of Theorem REF ,
the composition \(B_{2n+2} \overset{\bar{f}}{\longrightarrow } \operatorname{Sp}_{2n}({\mathbb {Z}}) \longrightarrow \operatorname{Sp}_{2n}({\mathbb {F}}_2)\) 
factors through \({{\mathfrak {S}}}_{2n+2}\) . Hence we have the inclusion \(\bar{f}(P_{2n+2}) \subset \operatorname{Sp}_{2n}({\mathbb {Z}})[2]\) .
The induced morphism \({{\mathfrak {S}}}_{2n+2} \rightarrow \operatorname{Sp}_{2n}({\mathbb {F}}_2)\) 
(which is non-trivial since the image of \(\sigma _1\)  in \(\operatorname{Sp}_{2n}({\mathbb {F}}_2)\)  is not the identity)
is injective as \({{\mathfrak {S}}}_{2n+2}\)  has no non-commutative proper quotient when \(n\ge 2\) .
This implies that Property (ii) above follows from (i). Let us now prove the latter.
In order to prove the opposite inclusion \(\bar{f}(P_{2n+2}) \supset \operatorname{Sp}_{2n}({\mathbb {Z}})[2]\) ,
we appeal to [1]}. In loc. cit. A'Campo considers the monodromy representation of \(B_{2n+2}\) 
in the free \({\mathbb {Z}}\) -module \(V\)  with basis \(\lbrace \delta _i, \, i=1 ,\ldots ,2n+1\rbrace \) ,
endowed with an alternating form \(I\)  satisfying \(I(\delta _i,\delta _{i+1})=1\)  for \(i=1,\ldots ,2n\) .
This form is degenerate with kernel \(K\)  spanned by \(\delta _1+\delta _3+\cdots +\delta _{2n+1}\) .
The monodromy representation maps each generator \(\sigma _i\)  of \(B_{2n+1}\) 
to the automorphism \(T_i\)  of \(V\)  defined by
\(T_i(\delta _j) ={\left\lbrace \begin{array}{ll}\delta _j& \text{ if } j\ne i-1,i+1,\\\delta _j+\delta _i &\text{ if }j=i-1,\\\delta _j-\delta _i &\text{ if }j=i+1.\end{array}\right.}\) 
Dividing by \(K\)  one obtains a representation of \(B_{2n+1}\)  in the symplectic group \(\operatorname{Sp}_{2n}({\mathbb {Z}})\) .
Now by [1]} the image of this monodromy representation contains the congruence subgroup \(\operatorname{Sp}_{2n}({\mathbb {Z}})[2]\) .
To conclude, it suffices to check that this representation is isomorphic to our \(\bar{f}\) . Indeed,
If we take
\(\left\lbrace  \delta _1,\delta _1+\delta _3,\ldots ,\delta _1+\delta _3+\cdots +\delta _{2n+1}, \delta _2,\delta _4,\ldots ,\delta _{2n} \right\rbrace \) 
as a basis of \(V\) , the matrix of the form \(I\)  becomes in \(V/K \cong {\mathbb {Z}}^{2n}\) 
the matrix \(J_{2n}\)  considered in § REF .
One easily checks that in this new basis the matrix of \(T_i\)  is equal to that of \(\bar{f}(\sigma _i)\)  for \(i=1,\ldots ,2n+1\) 
(for \(i=2n+1\)  one has to work modulo \(K\) , that is \(\delta _1+\delta _3+\cdots +\delta _{2n+1} = 0\) ).
